Advanced Higher T i m e : 3 h o u r s

MathematicsSpecimen Question Paper

NATIONALQUALIFICATIONS

©

Read carefully

1. Calculators may be used in this paper.

2. There are five Sections in this paper.

Section A assesses the compulsory units Mathematics 1 and 2Section B assesses the optional unit Mathematics 3Section C assesses the optional unit Statistics 1Section D assesses the optional unit Numerical Analysis 1Section E assesses the optional unit Mechanics 1.

Candidates must attempt all questions in Section A (Mathematics 1 and 2) and one of thefollowing:

Section B (Mathematics 3)Section C (Statistics 1)Section D (Numerical Analysis 1)Section E (Mechanics 1).

3. Full credit will be given only where the solution contains appropriate working.

[CO56/SQP182] 1

[C056/SQP182]

Marks

2

4

2

3

5

5

2

2

1

Section A (Mathematics 1 and 2)

All candidates should attempt this Section.

A1. (a) Find partial fractions for

(b) By using (a) obtain

A2. The performance of a prototype surface-to-air missile was measured on a

horizontal test bed at the firing range and it was found that, until its fuel was

exhausted, its acceleration (measured in m s–2) t seconds after firing was given

by

(a) Obtain a formula for its speed, t seconds after firing.

(b) The missile contained enough fuel for 10 seconds. What horizontal

distance would it have covered on the firing range when its fuel was

exhausted?

A3. Use the substitution x = 4 sin t to evaluate the definite integral

A4. Use Gaussian elimination to solve the system of linear equations

A5. (a) Find the derivative of y with respect to x, where y is defined as an

implicit function of x by the equation

(b) A curve is defined by the parametric equations

x = 2t + 1, y = 2t(t – 1).

(i) Find in terms of t.

(ii) Eliminate t to find y in terms of x.

Page two

4

42x −.

x

xdx

2

2 4−∫ .

a t t= + −8 10 3

42.

x

xdx+

−∫ 1

16 20

2

.

x y z

x y z

x y z

+ + =

− + = − ⋅

+ + = ⋅

0

2 1 1

3 2 0 9.

x xy y2 2 1+ + = .

dydx

[CO56/SQP182] 2

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5

1

1

3

4

1

2

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1

1

4

A6. Let u1, u2, ..., un, ... be an arithmetic sequence and v1, v2, ... , vn, ... be a

geometric sequence. The first terms u1 and v1 are both equal to 45, and the

third terms u3 and v3 are both equal to 5.

(a) Find u11.

(b) Given that v1, v2, ... is a sequence of positive numbers, calculate

A7. Use induction to prove that

for all positive integers n.

A8. Let the function f be given by

(a) The graph of y = f(x) crosses the y-axis at (0, a). State the value of a.

(b) For the graph of f(x)

(i) write down the equation of the vertical asymptote,

(ii) show algebraically that there is a non-vertical asymptote and state

its equation.

(c) Find the coordinates and nature of the stationary point of f(x).

(d) Show that f(x) = 0 has a root in the interval –2 < x < 0.

(e) Sketch the graph of y = f(x). (You must include on your sketch the

results obtained in the first four parts of this question.)

A9. Let z = cos θ + i sin θ.

(a) Use the binomial theorem to show that the real part of z4 is

Obtain a similar expression for the imaginary part of z4 in terms of θ.

(b) Use de Moivre’s theorem to write down an expression for z4 in terms

of 4θ.

(c) Use your answers to (a) and (b) to express cos 4θ in terms of cos θ and

sin θ.

(d) Hence show that cos 4θ can be written in the form k(cosmθ – cosnθ ) + pwhere k, m, n, p are integers. State the values of k, m, n, p.

Page three

vn

n =

∞

∑1

.

r r n n n

r

n

( ) ( )( )+ = + +=∑ 1 1

31 2

1

f x x x x

xx( )

( ), .= − + +

−≠2 7 4 5

22

3 2

2

cos cos sin sin .4 2 2 46θ θ θ θ− +

[CO56/SQP182] 3

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1

2

A10. In a chemical reaction, two substances X and Y combine to form a third

substance Z. Let Q(t) denote the number of grams of Z formed t minutes

after the reaction begins. The rate at which Q(t) varies is governed by the

differential equation

(a) Express in partial fractions.

(b) Use your answer to (a) to show that the general solution of the differential

equation can be written in the form

where A and C are constants.

State the value of A and, given that Q(0) = 0, find the value of C.

Find, correct to two decimal places,

(i) the time taken to form 5 grams of Z,

(ii) the number of grams of Z formed 45 minutes after the reaction

begins.

[END OF SECTION A]

Candidates should now attempt ONE of the following

Section B (Mathematics 3) on Page five

Section C (Statistics 1) on Page seven

Section D (Numerical Analysis 1) on Page nine

Section E (Mechanics 1) on Page eleven.

Page four

dQdt

Q Q= − −( )( ).

30 15900

90030 15( )( )− −Q Q

A Q

Qt Cln ,

3015

−−

= +( )

[CO56/SQP182] 4

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2

2

3

2

3

3

1

Section B (Mathematics 3)

ONLY candidates doing the course Mathematics 1, 2 and 3 should attempt this Section.

B11. Use the Euclidean Algorithm to find integers of x, y such that

195x + 239y = 1.

B12. The n × n matrix A satisfies the equation

A2 = 5A + 3I

where I is the n × n identity matrix.

Show that A is invertible and express A–1 in the form of pA + qI.

Obtain a similar expression for A4.

B13. Use Maclaurin’s theorem to write down the expansions, as far as the term in

x3, of

(i) where –1 < x < 1, and

(ii) where –1 < x < 1.

B14. Find the general solution of the differential equation

in each of the cases

(i) f(x) = 20cosx

(ii) f(x) = 20sin x

(iii) f(x) = 20cosx + 20sinx.

Page five

1+ x,

( ) ,1 2− −x

d ydx

dydx

y f x2

25 6− + = ( )

[CO56/SQP182] 5

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6

5

B15. (a) Show that the lines

intersect, and find the point of intersection.

(b) Let A, B, C be the points (2, 1, 0), (3, 3, –1), (5, 0, 2) respectively.

Find

Hence, or otherwise, obtain the equation of the plane containing A, Band C.

[END OF SECTION B]

Page six

L x y z

L x y z

1

2

32

1

36

1

31

6

211

2

:

:

− = + = −

−−

= − = −

→×

→AB AC.

[CO56/SQP182] 6

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2

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6

Section C (Statistics 1)

ONLY candidates doing the course Mathematics 1 and 2 and Statistics 1 should attempt this Section.

C11. An electronics company receives 65% of its components of a particular type

from supplier A and the remainder from supplier B. Of the components

supplied by A, 2% are defective while 5% of those supplied by B are

defective.

Calculate the probability that a randomly selected component, found to be

defective, was supplied by A.

C12. Records show a retailer that the number of orders for Noburn toasters during

any 28-day shopping period has a Poisson distribution with mean 8.

(a) Obtain the probability that, during a randomly selected 28-day shopping

period, the number of Noburn toasters ordered exceeds 8.

(b) State the number of Noburn toasters which should be in stock at the

beginning of a randomly selected 28-day shopping period in order to

ensure, with at least 95% probability, that demand during the period is

met from that stock.

C13. Find the probability of obtaining at least 55 heads in 100 tosses of a normal

coin.

C14. Given a list of all secondary schools in Scotland, outline the steps involved in

taking a cluster sample of Higher Mathematics pupils in 5th year.

State two reasons why it might be impractical to take a simple random sample

from the target population in this case.

C15. Of 40 fish caught in a loch, 15 were of a certain species.

Find an approximate 95% confidence interval for the proportion of this

species in the loch.

Explain what is meant by a 95% confidence interval.

State an assumption that you have made about the 40 fish.

Page seven[CO56/SQP182] 7

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C16. The management team at a large hospital determined that the time taken, in

minutes, to admit a patient was N(21,42).

(a) Calculate the probability that the mean time to admit a random sample of

10 patients is less than 17.7 minutes.

A new computer system for dealing with patient records was installed at the

hospital and the management team was interested in whether any reduction in

the mean time to admit patients had resulted. Following the introduction of

the system, a random sample of times (minutes) was as follows.

(b) Conduct an appropriate statistical test to evaluate the evidence for a

reduction in the mean waiting time.

[END OF SECTION C]

Page eight

23 20 16 10 20 12 23 18 16 19

[CO56/SQP182] 8

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Section D (Numerical Analysis 1)

ONLY candidates doing the course Mathematics 1 and 2 and Numerical Analysis 1 should attempt this Section.

D11. Obtain the Taylor polynomial of degree two for the function

Estimate the value of f(1.05) using the second degree approximation.

Write down the expression for the principal truncation error in this estimate,

and calculate its value. Compare the value of the principal truncation error

with the actual error in the approximation.

D12. The following data are available for function f.

Use the Lagrange interpolation formula to obtain a simplified quadratic

expression for f(x), using the values of f(x), where x = 1, 2 and 4.

D13. Derive the Newton forward difference formula of degree two.

The following table shows values which have been obtained for a function f.

Construct a difference table for f as far as the second differences and use the

Newton forward difference formula of degree two to obtain an approximation

to f(1.63).

Page nine

f x x x x( ) ln ( – ), , .= > near 3 2 2

3

ix

f(x)

0 1 2 3 4

1.6 1.7 1.8 1.9 2.0

0.826 1.203 1.609 2.042 2.503

xf(x)

1

2.7483

2

2.3416

3

1.8409

4

1.2268

[CO56/SQP182] 9

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6

D14. Use Simpson’s rule with two strips and the composite Simpson’s rule with

four strips to obtain two estimates, I1 and I2 respectively, for the integral

Perform the calculations using five decimal places.

The graph of y = f (iv)(x) is shown in the diagram.

The only zero of f (v)(x) on the interval [1, 2] occurs when x = 1.285 and

f (iv)(1.285) = 1.903. Use this information to obtain an estimate of the

maximum truncation error in I2. Hence write down the value of I2 to a

suitable accuracy.

Establish Richardson’s formula to improve the accuracy of Simpson’s rule by

interval halving and use it to obtain an improved estimate I3 for I based on the

values of I1 and I2.

[END OF SECTION D]

Page ten

I x e dxx= −∫ 4 2

1

2

.

6

4

2

0

–2

–4

–6

1 2

3

[CO56/SQP182] 10

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7

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7

Section E (Mechanics 1)

ONLY candidates doing the course Mathematics 1 and 2 and Mechanics 1 should attempt this Section.

Where appropriate, candidates should take the magnitude of theacceleration due to gravity as 9.8ms–2.

E11. A curler wishes to hit an opponent’s stone. Find the minimum speed with

which the curler’s stone must be projected horizontally towards the

opponent’s stone for the curler to succeed, given that the coefficient of friction

between a curling stone and the level surface of the ice is 0.04 and that the

distance from the point of release of the curler’s stone to the opponent’s stone

is 28 metres.

E12. A car travelling at 12ms–1 starts to accelerate 40 metres before leaving a built-

up area. The magnitude of the acceleration of the car t seconds later is given

Find the time taken before the speed of the car first reaches 26ms–1 and how

far outside the built-up area the car will then be.

E13. A fishing boat sails from a harbour on a bearing of 060° at a constant speed of

26 kilometres per hour. A cargo-ship is sailing on a bearing of 030° at a

constant speed of 18 kilometres per hour. The fishing boat leaves the harbour

at noon, at which time the cargo ship is 10 kilometres due East of the harbour.

Calculate the time at which the two vessels will be nearest to each other.

E14. A cricketer can throw a ball a maximum range of 60 metres to the wicket-

keeper. Given that the ball is released from a point 1.5 metres above the

ground and caught at the same height, calculate the speed with which it is

thrown and the maximum height reached above ground level. (Air resistance

may be ignored.)

E15. A packing case of mass 40 kg is to be moved through a vertical distance of

2 metres by means of a ramp of length 6 metres. The coefficient of friction

between the packing case and the ramp is such that, if the packing case was

placed on the ramp, it would just begin to slide down the ramp. Draw a

diagram showing the forces acting on the packing case and calculate this

coefficient of friction.

Calculate also the least force which would just move the packing case up the

ramp, when the force is applied in the horizontal direction.

[END OF SECTION E]

[END OF QUESTION PAPER]

Page eleven

by m s1

313 2 2( ) .− −t

[CO56/SQP182] 11

[CO56/SQP182] 12

Advanced HigherMathematicsSpecimen Solutions

NATIONALQUALIFICATIONS

©

[CO56/SQP182] 13

[C056/SQP182]

Section A (Mathematics 1 and 2)

A1. (a)

(b)

A2. (a)

(b)

Marks

[2]

[4]

[2]

[3]

Page two

4

4

4

2 2 2 2

1

2

1

2

2x x xA

xB

x

x x

−=

− +=

−+

+

=−

−+

( )( )

xx

dxx

dx

x xdx

x x x c

2

2 241 4

4

1 1

2

1

2

2 2

−= +

−

= +−

−+

= + − − + +

∫∫

∫ln( ) ln( )

a t t

v t t dt

t t t c

t v c

v t t t

= + −

= + −

= + − +

= = ⇒ =

= + −

∫

8 10 34

8 10 34

8 5 14

0 0 0

8 5 14

2

2

2 3

2 3

;

s v dt t t t c

t s c

t s

= = + − + ′

= = ⇒ ′ =

∴ = = + =

∫ 4 53

116

0 0 0

10 4005000

3625 1441

2

3

2 3 4

;

, – when

[CO56/SQP182] 14

Page three

A3.

A4.

Marks

[5]

[5]

x

xdx x t

tt

t dt dxdt t

x t

t tt

dt x t

t dt

t t

+

−=

= +−

⇒ =

= ⇒ =

= + × = ⇒ =

= +

= − + = + +

∫

∫

∫

∫

1

164

4 1

16 164 4

0 0

4 1 4

42

6

4 1

4 2 3 4

20

2

20

6

0

6

0

6

0

6

sin

sin

sincos cos

;

( sin ) cos

cos

( sin )

[ cos ]

/

/

/

/

π

π

π

π

π

π66

1 059≈ ⋅

1 1 1 0

2 1 1 1 1

1 3 2 0 9

1 1 1 0

0 3 1 1 1

0 2 1 0 9

1 1 1 0

0 3 1 1 1

0 0 1 0 5

− − ⋅

⋅

− − − ⋅

⋅

− − − ⋅

⋅ ( )r r r3 3 23 2″ = +

( )

( )

r r r

r r r

2 2 1

3 3 1

2′ = −

′ = −

Hence z y

x

= ⋅ = ⋅ − ⋅ = ⋅

= − ⋅ − ⋅ = − ⋅

0 5 1 1 0 5 3 0 2

0 2 0 5 0 7

; ( )/ ;

[CO56/SQP182] 15

A5. (a)

(b) (i)

(ii)

A6. (a)

(b)

A7.

Page four

Marks

[2]

[2]

[1]

[2]

[3]

[5]

x xy y

x x dydx

y y dydx

dydx

x yx y

2 2 1

2 2 0

22

+ + =

+ + + =

= − ++

( )

x t y t t

dxdt

dydt t dy

dxt

dy

dtdxdt

= + = −

= = − ⇒ = = −

2 1 2 1

2 4 2 2 1

; ( )

;

t x y x x

x x

= − = − − −

= − −

12

1 11

21 1

12

1 3

( ) ( ) ( )

( )( )

u d u

d

d

u

3 1

11

2 5

2 5 45

20

45 10 20

155

= + =

= −

= −

= + −

= −

( )

45 5

45

167

2

1

3 1

1

3

12

r

r v

S

=

=

=−

=

since ... are positive,

n

n

n

r r r r n n

n n n n n

n n n n

r

n

r

n

= = × =

= × × × =

=

+ = + + + +

= + + + + +

= + + +

∑ ∑=

1 1 2 2

1 2 3 2

1

1 1 1 2

1 2 1 2

1 2 3

1

3

1

1

3

1

3

LHS

RHS

True for

Assume true for and consider

=1

+1

.

( ) ( ) ( )( )

( )( ) ( )( )

( )( )( )

Thus if true for then true for + 1.

Therefore since true for = 1, true for all 1.

n n

n n ≥

[CO56/SQP182] 16

A8.

(a)

(b) (i) x = 2

(ii) After division, the function can be expressed in quotient/remainder

form:

Thus the line y = 2x + 1 is a slant asymptote.

(c) From (b), Turning point when

The stationary point at (3, 8) is a minimum turning point.

(d)

Hence a root between –2 and 0.

(e)

Marks

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[2]

Page five

f x x x x

x( )

( )= − + +

−2 7 4 5

2

3 2

2

x y a= ⇒ = ⇒ =0 54

54

f x x

x( )

( )= + +

−2 1 1

2 2

′ = −

−f x

x( )

( ).2 2

2 3

22

20

2 1

2 1 3

6

20

3

3

4

−−

=

− =

− = ⇒ =

′′ =−

>

( )

( )

( )( )

.

x

x

x x

f xx

x

for all

f f( )

( ); ( ) .− = − − − +

−< = >2 16 28 8 5

40 0 5

40

2

y = 2x + 1

x = 2

(3, 8)

[CO56/SQP182] 17

A9. (a) z4 = (cos θ + i sin θ)4

= cos4θ + 4 cos3θ (i sin θ ) + 6 cos2θ (i sin θ )2 + 4 cosθ (i sin θ )3 + (i sin θ )4

= cos4θ + 4i cos3θ sin θ – 6 cos2θ sin2 θ – 4i cosθ sin3 θ + sin4θ

= cos4θ – 6 cos2θ sin2θ + sin4θ + i (4cos3θ sinθ – 4 cosθ sin3θ)

Hence the real part is cos4 θ – 6 cos2θ sin2 θ + sin4 θ .

The imaginary part is (4 cos3θ sin θ – 4 cosθ sin3θ) = 4 cosθ sin θ (cos2θ – sin2θ)

(b) (cos θ + i sinθ )4 = cos 4θ + i sin 4θ

(c) cos 4θ = cos4θ – 6 cos2θ sin2θ + sin4 θ.

(d) cos 4θ = cos4θ – 6 cos2θ sin2θ + sin4 θ

= cos4θ – 6 cos2θ (1 – cos2θ ) + (1 – cos2θ )2

= cos4θ – 6 cos2θ + 6 cos4θ + 6 cos4θ + 1 – 2 cos2θ + cos4θ

= 8 cos4θ – 8 cos2θ + 1

= 8 (cos4θ – cos2θ ) + 1

ie k = 8, m = 4, n = 2, p = 1.

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Page six[CO56/SQP182] 18

A10. (a) 900 = A(15 – Q) + B(30 – Q)

Letting Q = 30 gives A = –60

and Q = 15 gives B = 60

(b)

(i)

(ii)

Marks

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Page seven

90030 15

6030

6015( )( ) ( ) ( )− −

= −−

+−Q Q Q Q

dQdt

Q Q

Q QdQ dt

Q QdQ dt

Q Q t C

QQ

t C

A

C

= − −

∴− −

=

∴ −−

+−

=

= +

−−

= +

=

= = ⋅

∫∫

∫ ∫

( )( )

( )( )

( ) ( )

ln ( – ) – ln ( – )

ln

30 15

900

90030 15

6030

6015

60 30 60 15

3015

60

60 2 41 59

ie 60

to 2 decimal places

ln

tQQ

QQ

Q t

=

=

= ⋅

6030

1560 2 60

30

2 15

5 60 2520

13 39

ln–

–– ln ln

ln

–

( – )

When = , = minutes to 2 decimal places.

ln( )

( )

( ) ( )

( )

,

/

/ /

/

/

30

2 15 60

30 2 15

2 1 30 1

30 1

2 1

45 10 36

60

60 60

60

60

−−

=

− = −

− = −

= −−

= = ⋅

QQ

t

Q Q e

Q e e

Q ee

t Q

t

t t

t

t

When grams to 2 decimal places.

[CO56/SQP182] 19

Section B (Mathematics 3)

B11.

B12.

B13. (i)

(ii)

Marks

[5]

[2, 2]

[3]

[2]

Page eight

A A I A A I

A A I A A I

A A I I A I

A A A I

2 4 2

2 2

1

3

5

3

1 1

3

5 3 5 3

5 3 25 30 9

155 84

5

= + = +

∴ = = + +

− = = +

∴ = −−

( )

–

( )

( ) is invertible and

f x x f

x

f x x f

f x x f

f x x f

x x

( ) ( )

( )

( ) ( ) ( )

( ) ( ) ( ) –

( ) ( ) ( )

/

/

/

/

= + =

= +

′ = + ′ =

′′ = − + ′′ =

′′′ = + ′′′ =

∴ + ≈ + −

−

−

−

1 0 1

1

12

1 0 12

14

1 0 14

38

1 0 38

1 1 12

1 2

1 2

3 2

5 2

118

116

2 3x x+

f x x f

f x x f

f x x f

f x x f

x x x x

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( )

= − =

′ = − ′ =

′′ = − ′′ =

′′′ = − ′′′ =

∴ − ≈ + + +

−

−

−

−

−

1 0 1

2 1 0 2

6 1 0 6

24 1 0 24

1 1 2 3 4

2

3

4

5

2 2 3

239 1 195 44

195 4 44 19

44 2 19 6

19 3 6 1

1 19 3 6

19 3 44 2 19

7 195 4 44 3 44

7 195 31 239 195

38 195 31 239

195 239 1 38 31

= × +

= × +

= × +

= × +

= − ×

= − − ×

= × − × − ×

= × − −

= × − ×

+ = = = −

So

ie when and

( )

( )

( )

x y x y

[CO56/SQP182] 20

B14.

(i)

(ii)

(iii)

B15. (a) L1: x = 3 + 2s; y = –1 + 3s; z = 6 + sL2: x = 3 – t; y = 6 + 2t; z = 11 + 2t

∴ for x: 3 + 2s = 3 – t ⇒ t = –2s∴ for y: 3s – 1 = 6 + 2t

7λ = 7 ⇒ s = 1; t = –2

∴ L1: x = 5; y = 2; z = 6 + s = 7

∴ L2: x = 5; y = 2; z = 11 + 2t = 11 – 4 = 7

ie L1 and L2 intersect at (5, 2, 7)

(b)

Marks

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[3]

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[6]

[5]

Page nine

f x x a x b x

a x b x a x b x a x b x x

a b

a b a b

b b a

y Ae Be x xx x

( ) cos ; cos sin

– cos – sin sin – cos cos sin cos

;

cos sin

= = +

⇒ + + + =

− =

+ = ⇒ = −

− = ⇒ = − =

= + + −

20

5 5 6 6 20

5 5 20

5 5 0

10 20 2 2

2 22 3

P.I.

Solution

d ydx

dydx

y f x

m m

m m

y Ae Bex x

2

2

2

2 3

5 6

5 6 0

2 3

− + =

− + =

∴ = =

= +

( )

A.E.

or

C.F.

f x x c x d x

c d c d

c d c d

y Ae Be x xx x

( ) sin ; cos sin

cos sin

= = +

− = ⇒ =

+ = ⇒ = =

= + + +

20

5 5 0

5 5 20 2

2 22 3

P.I.

Solution

f x x x

y Ae Be xx x

( ) cos sin

cos

= +

= + +

20 20

42 3Solution

A B C

AB AC

AB AC

x y z k

k

x y z

( , , ); ( , , ); ( , , )

–

–

– –

( , , )

.

2 1 0 3 3 1 5 0 2

3

1 2 1

3 1 2

3 5 7

3 5 7

2 1 0 1

3 5 7 1

−

→=

→=

→×

→= =

− − =

⇒ =

− − =

i j k i j k

i j k

i j k

+2 – ; – +2

Equation of plane has form

Equation is

[CO56/SQP182] 21

Section C (Statistics 1)

C11.

C12.

C13.

C14. Take a random sample of all schools with Higher Mathematics candidates

and then select every 5th year pupil, taking Higher Mathematics, from those

schools.

A list of all Higher Mathematics candidates would not be available to us and

there would be a high cost in terms of time and money if we had to visit

widely scattered schools.

C15.

For every one hundred intervals calculated we would expect 95 of them to

capture the true value of p and 5 not to.

We must assume that the 40 fish constitute a random sample.

C16.

Reject H0 at 1% level ie there is strong evidence of a reduction

in waiting time.

Marks

[5]

[2]

[2]

[4]

[2]

[2]

[6]

[4]

[5]

Page ten

P(A) P(def|A) Bayes Th. P(A|def)

P(B) P(def|B)

= ⋅ = ⋅ = ⋅ × ⋅⋅ × ⋅ + ⋅ × ⋅

=

= ⋅ = ⋅

0 65 0 02 0 02 0 650 02 0 65 0 05 0 35

2661

0 35 0 05

P p P

P

( ) ( ) ( )

( )

X X X

X

> = − ≤ = ⋅ ≤ = ⋅

≤ = ⋅

8 1 8 0 407 12 0 936

13 0 966} 13 required

B B P P ( , ) ( ) ( )100 1

255

55 50

50 184

1

2≥ ≈ ≥− −

= ⋅Z

ˆ

ˆ %

p

q

n

= ⋅

= ⋅ ⋅ ± ⋅ ⋅ × ⋅

= = ⋅ → ⋅

0 375

0 625 95 0 375 1 960 375 0 625

40

40 0 225 0 525

C.I. is

X N X Z

X

X

~ P( < )=P < –

H H tail test

P(

21 4

1017 7 17 7 21

4

10

0 005

17 7 21 21 1

17 7 0 005 0 01

2

0 1

,

: :

)

⋅ ⋅

= ⋅

= ⋅ = < −

≤ ⋅ = ⋅ < ⋅

µ µ

[CO56/SQP182] 22

Section D (Numerical Analysis 1)

D11.

D12.

D13.

Marks

[3]

[2]

[2]

[4]

[5]

[4]

Page eleven

f x x f xx

f xx

f xx

p x p h f hf h f

h h h h

x

( ) ( ); ( ) ; ( )( )

( )( )

( ) ( ) ( ) ( ) ( )

= − ′ =−

′′ = −−

′′′ =−

= + = + ′ + ′′

= + − ⋅ = − ⋅

ln

2nd degree polynomial is

ln

At

3 2 3

3 2

9

3 2

54

3 2

1 1 12

1

1 3 4 5 3 4 5

2

3

2 2

2

2 2

== ⋅ = ⋅ ⋅ ≈ ⋅

=−

= ⋅

= ⋅ − ⋅ = ⋅

1 05 0 05 1 05 0 1388

3

54

3 20 0011

0 1388 1 15 0 00096

3

3

, ( )

! ( )

h f

h

so

Principal error

Actual error ln

Values are equal to 3 decimal places

L x x x x x x x

x x x x x x

x

( )( )( )

( )( )( )( )

( )( )( )

.

( ) ( ) ( )

= − −− −

⋅ + − −−

⋅ + − − ⋅

= − + ⋅ − − + ⋅ + − + ⋅

= − ⋅ − ⋅ + ⋅

2 41 3

2 74831 41 2

2 34161 23 2

1 2268

6 8 2 74833

5 4 2 34162

3 2 1 22686

0 0502 0 2560 3

2 2 2

2

05450545

Consider the quadratic through

Let equation be

Then

Thus

0

0 0

0

( , ), ( , ), ( , )

( ) ( )( )

; ;

( )(

x f x f x f

y A A x x A x x x x

f A f A Ah Af f f

h

f

h

y fx x

hf

x x x

0 0 1 1 2 2

1 0 2 0 1

1 0 1 22 1 0

2

2

0

2

00

0

2

2 2

= + − + − −

= = + =− +

=

= +−

+−

∆

∆−−

= + < <

= + +−

x

hf

x x ph p

y f p fp p

f

1

2

2

0

0

0 0

2

0

2

0 1

1

2

)

, ,

( )

∆

∆ ∆

Setting when gives

x f f f

p

f

i i i i∆ ∆2

1 6 0 826

1 7 1 203377

29

1 8 1 609406

27

1 9 2 042433

28

2 0 2 503461

0 030 1

0 3

1 63 0 826 0 3 0 3770 3 0 7

20 029 0 936

⋅ ⋅

⋅ ⋅

⋅ ⋅

⋅ ⋅

⋅ ⋅

= ⋅⋅ = ⋅

⋅ ≈ ⋅ + ⋅ × ⋅ + ⋅ ⋅ ⋅ = ⋅( ).(– )

[CO56/SQP182] 23

D14.Marks

[4]

[2]

[4]

[2]

Page twelve

x f x m mf x m mf x

I I

E

( ) ( ) ( )4 2

1 2

1 00 0 13534 1 0 13534 1 0 13534

1 25 0 20040 4 0 80160

1 50 0 25205 2 0 50410 4 1 00820

1 75 0 28322 4 1 13288

2 00 0 29305 1 0 29305 1 0 29305

2 86697 1 43659

1 436596

0 23943 2 8669712

0 23891

1

180

⋅ ⋅ ⋅ ⋅

⋅ ⋅ ⋅

⋅ ⋅ ⋅ ⋅

⋅ ⋅ ⋅

⋅ ⋅ ⋅ ⋅

⋅ ⋅

= ⋅ = ⋅ = ⋅ = ⋅

≤ × 00 25 1 903 4 13 10 0 000041

0 2389

2 2 16 1

2

16 2

4 5

2

4

4 6 4

2

4 6

⋅ × ⋅ = ⋅ × = ⋅

= ⋅

= + + + = + +

= + + +

×

−

Hence

With strips and width 2 , the Taylor series for an integral

approximated by Simpson's rule, with principal truncation

error is

Similarly, with 2 strips of width ,

Taking

I

n h

O h

I I C h D h I Ch

n h

I I Ch Dh

n n

n

.

( ),

( ) ( ) .... .... –( )

... – ( )

( )) ( ) ( )

( )

( )

− = − +

⇒ = + −

= ⋅ + − ⋅ = ⋅

1 15 16

1

15

0 238911

150 00052 0 23888

2

6

2 2

3

gives

and

I I I O h

I I I I

I

n n

n n n

[CO56/SQP182] 24

Section E (Mechanics 1)

E11.

E12.

Marks

[4]

[5]

Page thirteen

R mg

ma F R

ma mg

a g

=

= − = −

= − ⋅

= − ⋅

µ

0 04

0 04

a t

v t t c

v t c v t t

t t t t

t t t

st t

t

= −

= − +

= = ⇒ = ⇒ = − +

− + = ⇒ − + =

− − = =

= −

+ +

−

13

13 2

13

13

12 0 12 13

13 12

13

13 12 26 13 42 0

6 7 0 6 7

13

13

2 312

2

2

2 2

1

2 3

( )

( )

( )

( )

( )( ) ,

First reaches 26 m s after 6 secs.

cc

s t c

s t t t

t s

= − = ⇒ = −

= − + −

= = − + − =

40 0 40

136 9

12 40

6 78 24 72 40 86

2 3

m outside the built-up area.

v u as

u g

u u

2 2

2

2 1

2

0 2 0 04 28

21 952 4 7

= +

= − × ⋅ ×

= ⋅ ⇒ = ⋅ −m s

mg

R

F

[to 1 decimal place]

[CO56/SQP182] 25

E13.

E14.

Marks

[7]

[6]

Page fourteen

v i j i j

r i j r i j

r r i j

r r

r r

C

F c

F C

F C

F C

t t t t

t t

t t

d

dtt t

= + = + ⋅

= ⋅ + = + + ⋅

⇒ − = ⋅ − − ⋅

− = ⋅ − + ⋅

−= ⋅ − + ⋅

18 30 18 30 9 15 6

22 5 13 9 10 15 6

13 5 10 2 6

13 5 10 6 76

27 13 5 10 13 5

22 2

2

sin cos

( )

( )

( )

( )

==

⇒ =

= ⋅

0

378 270

0 71

for stationary point

h

= 43min

t

t

Ru

gug u

u

u gh

gh

h gg

= ⇒ = = ⇒ =

= ⋅

= −

= × −

= =

= ⋅

−

2 22

1

2 2

260 60

24 2

0 2

0 60 12

2

302

15

16 5

sin

sin

α

α

Max range g

m s

Max height when

g

m

So max height above ground m

60°

CF x0

y

30°

26 kmh–1 18 kmh–1

0F C10 km

[CO56/SQP182] 26

Vessels will be closest at 12.43 pm

E15.Marks

[3]

[7]

Page fifteen

Perp to plane

Parallel to plane in

On point of slipping

or 3 5)

R g

F g s

F R

g g

=

=

=

⇒ =

= = = = ⋅

40

40

40 40

2

32

2

4 2

2

4

cos

cos sin

tan (

α

α

µ

µ α α

µ α

Perp to plane

Parallel to plane in

N

R g P

F g s P

F R g P g P

g P g P

g P

P

= +

+ =

= ⇒ +

+ =

+ + =

= −

= × ⋅⋅ ×

= ⋅

40

40

2

4

2

440

32

6

1

340

1

3

32

6

403

212

403

2 23

80 2 14

2

80 9 8

1 75 2316 8

cos sin

cos

.

( )

α α

α αF

F

40 g

40 g

R

RP

6m

α

α

2m

(candidates likely to use µ = 0.35

and numerical value of α)

[END OF MARKING INSTRUCTIONS]

α

[CO56/SQP182] 27

[CO56/SQP182] 28